Question 5.8: Water flows out of the large tank and through the pipeline s......
Water flows out of the large tank and through the pipeline shown in Fig. 5–20. Draw the energy and hydraulic grade lines for the pipe if friction losses are neglected.
Fluid Description. We will assume the water level in the tank remains essentially constant, so that V_A = 0 and steady flow will be maintained. The water is assumed to be an ideal fluid.
Energy Grade Line. We will establish the datum through DE. At A the velocity and pressure heads are both zero, and so the total head is equal to the elevation head, which is at a level of
The EGL remains at this level since the fluid is ideal, and so there are no energy losses due to friction.
Notice that if the Bernoulli equation is applied at points A and B, then V_B = 0 because we have assumed V_A = 0. This result, although not quite correct, requires the fluid to accelerate only within the pipe, from rest at B to a velocity V_{B^{\prime}} \text{at} B^{\prime}. Here we will assume these two points are rather close together, although actually they are somewhat separated. See Sec. 9.6.
Hydraulic Grade Line. Since the (gage) pressure at both A and E is zero, the velocity of the water exiting the pipe at E can be determined by applying the Bernoulli equation on a streamline that contains these two points.
\frac{p_A}{\gamma} + \frac{V_A^2}{2g} + z_A = \frac{p_E}{\gamma} + \frac{V_E^2}{2g} + z_E0 + 0 + 9 m = 0 + \frac{V_E^2}{2(9.81 m/s^2)} + 0
V_E = 13.29 m/s
The velocity of the water through pipe section B′C can now be determined from the continuity equation, considering the fixed control volume to contain the water within the entire pipe. We have
0 – V_B′ A_B′ + V_E A_E = 0
– V_{B′}[\pi (0.1 m)^2] + (13.29 m/s)[\pi (0.05 m)^2] = 0
V_{B′} = 3.322 m/s
The HGL can now be established. It is located below the EGL, a distance defined by the velocity head V^2/2g. For pipe segment B′C this head is
\frac{V^2_{B^{\prime}}}{2g} = \frac{(3.322 m/s)^2}{2(9.81 m/s^2)} = 0.5625 mThe HGL is maintained at 9 m – 0.5625 m = 8.44 m until the transition at C changes the velocity head along CDE to
\frac{V_{E}^2}{2g} = \frac{(13.29 m/s)^2}{2(9.81 m/s^2)} = 9 mThis causes the HGL to drop to 9 m – 9 m = 0. Along CD, z is always positive, Fig. 5–20, and therefore, a corresponding negative pressure head -p/\gamma must be developed within the flow to maintain a zero hydraulic head, i.e., p/\gamma + z = 0. If this negative pressure becomes large enough, it can cause cavitation, something we will discuss in the next example.
Finally, along DE both p/\gamma and z are zero.